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Arada [10]
2 years ago
7

Calculate the percent activity of the radioactive isotope strontium-89 remaining after 5 half-lives.

Chemistry
1 answer:
maria [59]2 years ago
5 0
The answer to this question would be: 3.125%

Half-life is the time needed for a radioactive molecule to decay half of its mass. In this case, the strontium-89 is already gone past 5 half lives. Then, the percentage of the mass left after 5 half-lives should be:
100%*(1/2^5)= 100%/32=3..125%
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3 0
1 year ago
Help me please............…
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In the following reaction, how many grams of oxygen will react with 10.47 grams of benzene (C6H6)? 2C6H6 + 1502 12CO2 + 6H2O The
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21 mL It required 42.35 mL of H2SO4 to neutralize 21.17 mL of 0.5000 M NaOH. Calculate the concentration of H2SO4
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The balanced equation for the above reaction is 
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of NaOH moles required-0.5000 M / 1000 mL/L x 21.17 mL = 0.010585 mol
According to stoichiometry, acid moles required are 1/2 of the base moles reacted
Therefore number of H₂SO₄ moles reacted - 0.010585 /2  mol
Number of moles in 42.35 mL of H₂SO₄ - 0.010585 /2 mol
Therefore in 1 L solution - (0.010585) /2 / 42.35 mL x 1000 mL/L = 0.125 M
Molarity of H₂SO₄ - 0.125 M
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